Two objects of masses m 1 = 0.40 kg and m 2 = 0.96 kg are placed on a horizontal

Question

Two objects of masses

m1 = 0.40 kg

and

m2 = 0.96 kg

are placed on a horizontal frictionless surface and a compressed spring of force constant

k = 250 N/m

is placed between them as in figure (a). Neglect the mass of the spring. The spring is not attached to either object and is compressed a distance of 9.4 cm. If the objects are released from rest, find the final velocity of each object as shown in figure (b). (Let the positive direction be to the right. Indicate the direction with the sign of your answer.)

Explanation / Answer

conservation of momentum

pf = pi

m1v1 + m2v2 = 0

v2 = -(m1/m2)v1 ...(1)

since the surface is frictionless , conservation of energy

final kinectic energy of both blocks is zero

1/2 * m1v1^2 + 1/2 m2v2^2 = 1/2 kx^2 ...(2)

putting the value of v2 from equation 1 in eq (2)

m1(1+(m1/m2)v1^2 = kd^2

v1 = -d*sqrt(k/m1(1+m1/m2)

given : d = 9.4 cm = 9.4 x 10^-2 m

m1 = 0.4 kg

m2 = 0.96 kg

k = 250 N/m

v1 = -1.97 m/s

here negative signs shows that m1 block is moving to the left

now put the value of v1 in eq (1)

v2 = 0.82 m/s to the right

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